A phosphate buffer solution was made by adding 22 g of Na2HPO4 (FW 142g/mol) and

Question

A phosphate buffer solution was made by adding 22 g of Na2HPO4 (FW 142g/mol) and 6.2 g of NaH2PO4. H2O (FW 138 g/mol) to enough water to make a 1 L solution. The pKa of the buffer is 6.86.

a. What is the pH of the buffer solution? Final answer in 2 sig figs.

b. You take 150mL of this buffer and add 1mL of a 5 M NaOH solution. what is the pH of the resulting solution?

c. Now you take another 150mL of original buffer (from part a) and add 2.5mL of conc. HCL solution (12M). What is the pH of the resulting solution?

Explanation / Answer

Answer:

a) The buffer formed is  NaH2PO4 - Na2HPO4.

Let us calculate number of moles and molarity of each species forming buffer.

# of moles of NaH2PO4 = Given mass / molar mass = 6.2 / 138 = 0.045

hence, [NaH2PO4] = 0.045/ 1L = 0.045 M/L

# of Moles of Na2HPO4 = 22 / 142 = 0.155

hence, [Na2HPO4] = 0.155 / 1L = 0.155 M/L

pKa2 for H3PO4 = Pka for NaHH2PO4 = 7.21

Henderson Hasselbalch equation give pH of the acidic buffer as,

pH = pKa + log([conjugate base]/[Acid]).

For above buffer,

pH = pKa + log([Na2HPO4]/[NaH2PO4])

pH = 7.21 + log(0.155/0.045)

pH = 7.21 + 0.54

pH = 7.75

pH of original buffer = 7.75.

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b) pH of 150mL (0.150 L) of this buffer and add 1mL (10-3 L) of a 5 M NaOH solution.

Let us calculate # of moles of NaoH adde to buffer,

# of moles of NaOH = Molarity x Volume in L = 5 x 10-3 = 0.005 moles.

Let us calulate original # of moles of Na2HPO4 and NaH2PO4 .

Original # of moles of Na2HPO4   = 0.155 x 0.150 = 0.023 moles.

Original # of moles of NaH2PO4 = 0.045 x 0.150 = 0.007 moles.

NaOH is a strong base and hence on addition to the buffer it will lower the mole number of Acidic specie i.e.NaH2PO4 and will increase # of moles of conjugate base i.e Na2HPO4.

New # of moles of Na2HPO4   = 0.023 + 0.005 = 0.028 moles.

New # of moles of NaH2PO4 = 0.007 - 0.005 = 0.002 moles.

Using these values in Henderson - Hasselbalch equation we get,

pH = 7.21 + log(0.028/0.002) ............. (being volume same number of moles used as concentration ratio Mathematically same)

pH = 7.21 + 1.15

pH = 8.36

pH of NaOH added buffer = 8.36.

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c) Addition of 2.5mL of conc. HCL solution (12M).

# of moles of HCl addded = 12 x 0.0025 = 0.03 moles

Addition of strong acid will increase # of moles of Acid componant in buffer and will lower the concentration of basic componant.

New number of moles of Na2HPO4 = 0.028 - 0.03 = -0.02

Techanically there will be no basic componant left in buffer and now there is excess of H+ ions in the solution.

Hence the solution exceeds it's buffer capacity.

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